PD Control Example

A  common control scheme is called PID or Proportional-Derivative-Integral control. PID controller are used in everything from process controls to motor controls. The following example illustrates the use of a PD, Proportional-Derivative, control or a rotating mass.

6.1 Physical Model

A simple DC motor model is shown below. It consists of a rotating mass, an input torque and viscous friction.

where J=10
and b=20
The input is torque (T) and the output is motor position (θ)

The system transfer can be found as follows:

substituting in value for b and J gives and the transfer function is

We wish to control the motor position (θ) and have a damping ratio (ζ) of 0.707

6.2 Open-Loop Analysis

The open characteristics of the system, shown below, don't meet the design specification. First the system does not have a ζ of 0.707. Second, the output is an integral of the input (pole at the origin). We wanted the output to follow the input.

6.2 Proportional Control

Use proportional feedback to control the motor position. Proportional control has the form

Substituting into the differential equation for the system gives

This can also be done using block diagrams and transfer functions

where θ is the measured position, θ_d is the desired position and k_p is the feedback gain.

Now simplify the transfer function

Note that when analyzing the closed loop system using root locus the characteristic equation is

We can find k_p using either root locus or, if we know the desired closed-loop poles, we can find k_p directly through algebra.

Finding k_p Using Root Locus

Given the closed loop characteristic equation

plot the poles (roots of d(s)) and zeros (roots of n(s)) as 0 ≤ k_p ≤ ∞.

From all of the possible solutions choose k_p that results in ζ=0.707. Do this using brute force by selecting a k_p and finding the roots of the resulting quadratic equation. As you change k_p the root locations are indicated with an x on the root locus plot.

Solving for k_p directly

If you know the form of the characteristic equation for the closed-loop system, you can often solve directly for k_p. In our problem closed-loop characteristic equation is:

Compare this to the standard 2^nd order form

For the two equations to be equal each terms in the two equations must match. Set ζ=0.707, match terms and solve:

Solve for k_p

Alternately if we know the desired close-loop pole locations we can solve for k_p directly. The key is that you must know the exact value for the pole locations and those must lie on the root locus. The desired pole locations from the root-locus graph are: s= -0.025±-0.025 i. Therefore the desired closed-loop equation must be

The actual closed-loop characteristic equation is matching terms with the closed-loop equation yields

Verifying the Solution

Use the following simulation to verify the solution. Select a value for k and change the value for θ_desired. The settling time and damping ratio for the system should match the calculate settling time (remind me). Experiment with different value for k and verify that you calculated closed-loop poles (T_settling and ζ) match the simulated response.

In addition to tracking θ_desired the system also rejects disturbances to the motor. Click and drag on the wheel. Note how the torque increases to resist the disturbance.

The final check in the design is to verify that the motor torque requirements don't exceed the physical limits of the system. Starting at 0° set θ_desired to 90° (or some value that represents the maximum change in position) and observe the maximum torque required by the control system. If this value exceeds the physical limits of the system you may need to reduce k_p, limit the maximum change in the system, or limit how fast the system is allowed to change (this technique is discussed later in this text).